IIT JEE , AIEEE Forum - Physics , Mechanics

garima4

A sphere starts rolling down an incline of inclination theta.Find the speed of its center when it has covered a distance L ?

iit009

is answer is square root 10/7gLsin(theta)

garima4

Yes u r correct but please explain me all the steps involved in it.

ashish_banga

a cceleration of the sphere is given by = g sintheta/ 1 + I / mr^2
I= moment of inertia of the sphere about its centre
use V^2 = 2 a L

SCIENTIST135

distance = L let t be the time taken. speed = distance / time = L / t

as initial velocity is zero, u = 0, S = ut +at²/2 = 0 + gsin@.t² / 2

S = gsin@.t²/2 , hence t²= 2S/gsin@ or t = sqrt(2S/gsin@) let @ = theta

as S = L , t = sqrt(2L/gsin@) . now,

now speed = L/t = L / [sqrt(2L/gsin@)] = sqrt ( Lgsin@ / 2)

speed = sqrt ( Lgsin@ / 2).

SCIENTIST135

you can consider spherical body as a point object in this case.

me_coolguy_iit_aspirant

hey u cant say .{distance =speed*time}
wen a body experiences acceleration.

me_coolguy_iit_aspirant

it is because the body is experiencing accn. of [gsin

]in the direction of its motion....

SCIENTIST135

OOPS IM SORRY. YES. THEN WE SHOULD USE v^2 = u^2 + 2aL
v^2 = 2.gsin@.L
v = sqrt (2Lgsin@)

a5hw1n_5

let a be the acceleration along the plane.
balancing the forces along the incline
mgsin

- f = ma -------1
where f is frictional force
balancing torque about the centre
fR= Ie where e is the angular accleration
e= a/R and I = 2/5mR² (for a sphere)
hence
fR= 2/5mR²a/R
f= 2/5ma
sub in eq 1
mgsin

-2/5ma= ma
gsin

= 7/5a
a= 5/7gsin

we know that v =

2La
sub a to get the ans

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